Răspuns :
2^(n+1)=2^n*2
3*2^(n+2)=3*2^n*4=12*2^n
insumezi si obtii 2^n*2+2^n*5+2^n*12 = 2^n(2+5+12)=19*2^n, divizibil cu 19 deoarece 19 este divizibil cu 19
3*2^(n+2)=3*2^n*4=12*2^n
insumezi si obtii 2^n*2+2^n*5+2^n*12 = 2^n(2+5+12)=19*2^n, divizibil cu 19 deoarece 19 este divizibil cu 19
2^(n+1) + 5*2^n + 3*2^(n+2)
[tex]\it \ a=2^{n+1} +5\cdot2^n+3\cdot2^{n+2} =2^n\cdot2+5\cdot2^n+ 3\cdot2^n\cdot2^2 = \\\;\\ = 2^n(2+5+12) = 2^n\cdot19 \in M_{19} \Longrightarrow a \vdots 19[/tex]
[tex]\it \ a=2^{n+1} +5\cdot2^n+3\cdot2^{n+2} =2^n\cdot2+5\cdot2^n+ 3\cdot2^n\cdot2^2 = \\\;\\ = 2^n(2+5+12) = 2^n\cdot19 \in M_{19} \Longrightarrow a \vdots 19[/tex]
